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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 241-270 of 337 questions

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241 Medium
241 of 337
Application +4 | -1
Let p,q∈Rp, q \in \mathbb{R} and (1−3i)200=2199(p+iq),i=−1(1-\sqrt{3} i)^{200}=2^{199}(p+i q), i=\sqrt{-1}, then p+q+q2p+q+q^{2} and p−q+q2p-q+q^{2} are roots of the equation.
View answer and explanation
Correct option: A
x2−4x+1=0x^{2}-4 x+1=0
The correct answer is A (x2−4x+1=0x^{{2}} - 4x + 1 = 0) because it is the quadratic equation whose roots are derived from the given complex number equation.
242 Medium
242 of 337
Analyzing +4 | -1
The imaginary part of (z−1)(cos⁡α−isin⁡α)+(z−1)−1×(cos⁡α+isin⁡α)\left( {z - 1} \right)(\cos \alpha - i\sin \alpha ) + {\left( {z - 1} \right)^{ - 1}} \times (\cos \alpha + i\sin \alpha ) zero, if
View answer and explanation
Correct option: C
arg⁡(z−1)=α\arg \left( {z - 1} \right) = \mathrm{\alpha }
Option C (arg⁡(z−1)=α\arg \left( {z - 1} \right) = \mathrm{\alpha }) is correct because the imaginary part of the expression is zero when arg⁡(z−1)=α\arg(z-1) = \alpha.
243 Medium
243 of 337
Application +4 | -1
The value of ∑r=18(sin⁡2  r  π9+icos⁡2  r  π9),\sum \limits_{r = 1}^8 \left( {\sin \frac{{2\;r\;\pi }}{9} + i\cos \frac{{2\;r\;\pi }}{9}} \right), is
View answer and explanation
Correct option: D
−i- i
Option D (−i- i) is correct because it is the simplified value of the given summation after applying the formula for the sum of a geometric series and using properties of complex numbers.
244 Medium
244 of 337
Application +4 | -1
If z=−12+i32z=\frac{-1}{2}+i \frac{\sqrt{3}}{2}, then 8+10z+7z28+10 z+7 z^2 is equal to
View answer and explanation
Correct option: C
−12+i332\frac{-1}{2}+i \frac{3 \sqrt{3}}{2}
Option A ((0,2 tan-1 1/4]) is correct because it is the interval for 'a' where the discriminant of the quadratic equation is greater than zero.
245 Medium
245 of 337
Application +4 | -1
If z=x+iyz = x + iy, then area of the triangle whose vertices are points z,  iz,  z+izz,\;iz,\;z + iz is
View answer and explanation
Correct option: A
12∣z∣2\frac{1}{2}{\left| z \right|^2}
Option A (12∣z∣2\frac{1}{2}{\left| z \right|^2}) is correct because the area of the triangle formed by the given vertices is half the square of the magnitude of z.
246 Medium
246 of 337
Application +4 | -1
If (1+cos⁡ϕ+isin⁡ϕ1+cos⁡ϕ−isin⁡ϕ)=u+iv\left( {\frac{{1 + \cos \phi + i\sin \phi }}{{1 + \cos \phi - i\sin \phi }}} \right) = u + iv, where uu and vv all real numbers, then uu is
View answer and explanation
Correct option: B
cos⁡nϕ\cos n\phi
Option B (cos⁡nϕ\cos n\phi) is correct because the expression simplifies to cos⁡nϕ+isin⁡nϕ\cos n\phi + i\sin n\phi, and thus u = cos⁡nϕ\cos n\phi.
247 Medium
247 of 337
Application +4 | -1
The centre of a regular hexagon is at the point z=iz = i. If one of its vertices is at 2+i2 + i, then the adjacent vertices of 2+i2 + i are at the points
View answer and explanation
Correct option: D
1+i(1±3)1 + i\left( {1 \pm \sqrt 3 } \right)
Option D (1+i(1±3)1 + i\left( {1 \pm \sqrt 3 } \right)) is correct because it represents the vertices adjacent to 2+i2+i in a regular hexagon centered at ii.
248 Medium
248 of 337
Application +4 | -1
If z=x−iyz=x-i y and z13=p+iqz^{\frac{1}{3}}=p+i q, then (xp+yq)/(p2+q2)\left(\frac{x}{p}+\frac{y}{q}\right) /\left(p^{2}+q^{2}\right) is equal to
View answer and explanation
Correct option: A
-2
Option A (-2) is correct because by cubing z13=p+iqz^{\frac{1}{3}} = p + iq and equating real and imaginary parts, the expression simplifies to -2.
249 Medium
249 of 337
Application +4 | -1
If z=x+iyz = x + iy is a variable complex number such that arg⁡z−1z+1=π4\arg \frac{{z - 1}}{{z + 1}} = \frac{\pi }{4}, then
View answer and explanation
Correct option: C
x2+y2−2y=1{x^2} + {y^2} - 2y = 1
Option C (x2+y2−2y=1{x^2} + {y^2} - 2y = 1) is correct because by substituting z=x+iyz = x + iy into the given argument equation and simplifying, we obtain the equation of a circle.
250 Medium
250 of 337
Application +4 | -1
If z1,z2\mathrm{z}_{1}, \mathrm{z}_{2} and z3,z4\mathrm{z}_{3}, \mathrm{z}_{4} are 2 pairs of complex conjugate numbers, then
arg⁡(z1z4)+arg⁡(z2z3)\arg \left(\frac{z_{1}}{z_{4}}\right)+\arg \left(\frac{z_{2}}{z_{3}}\right) equals:
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the arguments of a complex number and its conjugate is zero, and z2 and z4 are conjugates of z1 and z3 respectively.
251 Medium
251 of 337
Application +4 | -1
If n∈Nn \in \mathbf{N}, then (1+i)n(1−i)n−2\frac{(1+i)^n}{(1-i)^{n-2}} is equal to
View answer and explanation
Correct option: B
−2in+1-2 i^{n+1}
Option B (−2in+1-2 i^{n+1}) is correct because after simplifying the expression using properties of complex numbers, the result matches this option.
252 Medium
252 of 337
Application +4 | -1
If Z{Z} is a unimodular complex number then arg⁡(z2+zˉ)=\arg \left(z^2+\bar{z}\right)=
View answer and explanation
Correct option: A
12arg⁡z\frac{1}{2} \arg z
Option A (12arg⁡z\frac{1}{2} \arg z) is correct because after simplification using trigonometric identities, the argument of the expression reduces to half of the argument of z.
253 Medium
253 of 337
Application +4 | -1
The minimum value of ∣a+bω+cω2∣\left| {a + b\omega + c{\omega ^2}} \right|, where a,  ba,\;b and cc are all not equal integers and ω(≠1)\omega \left( { \ne 1} \right) is a cube root of unity, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the minimum value of the expression is 1 when a, b, and c are distinct integers.
254 Medium
254 of 337
Application +4 | -1
Let z1,z2,z3{z_1},{z_2},{z_3} be three vertices of an equilateral triangle circumscribing the circle ∣z∣=12.\left| z \right| = \frac{1}{2}. If z1=12+i32{z_1} = \frac{1}{2} + i\frac{{\sqrt 3 }}{2} and z1,z2,z3{z_1},{z_2},{z_3} were in anticlockwise sense, then z2{z_2} is
View answer and explanation
Correct option: D
−1- 1
The correct answer is D (−1- 1) because, in an equilateral triangle, the vertices are equally spaced around the circumscribing circle, and using the properties of complex numbers, z2 = -1.
255 Medium
255 of 337
Application +4 | -1
Principal Argument of complex number (1+i)5(1+3i)2−2i(−3+i)\frac{(1+i)^5(1+\sqrt{3} i)^2}{-2 i(-\sqrt{3}+i)} is
View answer and explanation
Correct option: C
−5π12\frac{-5 \pi}{12}
Option C (-5π/12) is correct because it is the principal argument obtained after calculating the argument of the complex number and adjusting to the range (-π, π].
256 Medium
256 of 337
Application +4 | -1
If a=cos⁡θ+isin⁡θ,  then  1+a1−aa = \cos \mathrm{\theta } + i\sin \mathrm{\theta },\mathrm{\;then\;}\frac{{1 + a}}{{1 - a}} is equal to
View answer and explanation
Correct option: C
icot⁡θ2i\cot \frac{\mathrm{\theta }}{2}
Option C (icot⁡θ2i\cot \frac{\mathrm{\theta }}{2}) is correct because it is the simplified form of the given expression using Euler's formula and trigonometric identities.
257 Medium
257 of 337
Application +4 | -1
Value of ∑k=16[sin⁡(2kπ7)−icos⁡(2kπ7)]\sum \limits_{k = 1}^6 \left[ {\sin \left( {\frac{{2k\pi }}{7}} \right) - i\cos \left( {\frac{{2k\pi }}{7}} \right)} \right] is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the summation simplifies to 'i', which is not among the given options.
258 Medium
258 of 337
Application +4 | -1
The principal value of the arg⁡(z)\arg (z) and ∣z∣|z| of the complex number z=(1+cos⁡θ+isin⁡θ)5(cos⁡θ+isin⁡θ)3z=\frac{(1+\cos \theta+i \sin \theta)^5}{(\cos \theta+i \sin \theta)^3} are
View answer and explanation
Correct option: A
−θ2,32cos⁡5θ2-\frac{\theta}{2}, 32 \cos ^5 \frac{\theta}{2}
Option A (-θ2\frac{{\theta}}{{2}}, 32cos⁡5θ232 \cos ^5 \frac{{\theta}}{{2}}) is correct because after simplifying using trigonometric identities and De Moivre's theorem, the argument and modulus are -θ2\frac{{\theta}}{{2}} and 32cos⁡5θ232 \cos ^5 \frac{{\theta}}{{2}}, respectively.
259 Medium
259 of 337
Application +4 | -1
If z=x+iyz=x+i y is a complex number with x,y∈Qx, y \in Q and ∣z∣=1|z|=1, then ∣z2n−1∣\left|z^{2 n}-1\right| is a _____ for every n∈Nn \in N.
View answer and explanation
Correct option: B
Rational number
Option B (Rational number) is correct because given the conditions, |z(2n) - 1| can be shown to be a rational number using trigonometric identities and properties of rational numbers.
260 Medium
260 of 337
Application +4 | -1
If z(z+α‾)+zˉ(z+α)=0z\left( {\overline {z + \alpha } } \right) + \bar z\left( {z + \alpha } \right) = 0 where α\alpha is a complex constant, then zz is represented by a point on
View answer and explanation
Correct option: A
A circle
Option A (A circle) is correct because the given equation can be rearranged into the standard form of a circle's equation in the complex plane.
261 Medium
261 of 337
Application +4 | -1
If z=−2+23iz=-2+2 \sqrt{3} i, then z2n+22nzn+24nz^{2 n}+2^{2 n} z^n+2^{4 n} may be equal to
View answer and explanation
Correct option: C
3⋅42n,n3 \cdot 4^{2 n}, n is multiple of 3
Option C (3⋅42n,n3 \cdot 4^{2n}, n is multiple of 3) is correct because when n is a multiple of 3, the expression simplifies to this value.
262 Medium
262 of 337
Application +4 | -1
The value of (sin⁡π8+icos⁡π8)8(sin⁡π8−icos⁡π8)8\frac{\left(\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}\right)^8}{\left(\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}\right)^8}
View answer and explanation
Correct option: C
1
Option C (1) is correct because after simplification using Euler's formula, the expression evaluates to 1.
263 Medium
263 of 337
Application +4 | -1
∑k=16(sin⁡(2kπ7)+icos⁡(2kπ7))=\sum\limits_{{k=1}}^6\left(\sin \left(\frac{2 k \pi}{7}\right)+i \cos \left(\frac{2 k \pi}{7}\right)\right)=
View answer and explanation
Correct option: C
−i-i
Option C (-i) is correct because by converting to exponential form and using geometric series formula, the sum simplifies to -i.
264 Medium
264 of 337
Analyzing +4 | -1
If z{z} satisfies the equation (z−2z+2)(zˉ−2zˉ+2)=1\left(\frac{z-2}{z+2}\right)\left(\frac{\bar{z}-2}{\bar{z}+2}\right)=1, then minimum value of ∣z∣|z| is equal to:
View answer and explanation
Correct option: A
0
The correct answer is A (0) because the given equation simplifies to |z-2| = |z+2|, which represents the perpendicular bisector of the line joining 2 and -2, i.e., the y-axis. The minimum value of |z| is 0, occurring at the origin.
265 Medium
265 of 337
Application +4 | -1
The multiplicative inverse of (6+5i)2(6+5 i)^2 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the multiplicative inverse of (6+5i)2(6+5i)^2 is 11(61)2−60(61)2i\frac{11}{(61)^2} - \frac{60}{(61)^2}i, which is not among the given options.
266 Medium
266 of 337
Application +4 | -1
If ∣z+4∣≤3|\mathrm{z}+4| \leq 3 then maximum and minimum value of ∣z+1∣|z+1| is .
View answer and explanation
Correct option: C
6,0
Option C (6,0) is correct because the maximum and minimum values of ∣z+1∣|z+1| are found by using the triangle inequality on ∣(z+4)−3∣|(z+4)-3|.
267 Medium
267 of 337
Application +4 | -1
In a geometrical progression, first term and common ratio are both 12(3+i)\frac{1}{2}(\sqrt{3}+i). Then the absolute value of the nnth term of the progression is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the absolute value of the nth term of the geometric progression is 1, as the magnitude of the common ratio is 1.
268 Medium
268 of 337
Application +4 | -1
The value of x4+9x3+35x2−x+4{x^4} + 9{x^3} + 35{x^2} - x + 4 for x=−5+2−4x = - 5 + 2\sqrt { - 4} is
View answer and explanation
Correct option: B
−160- 160
Option B (−160- 160) is correct because substituting x=−5+2−4x = - 5 + 2\sqrt { - 4} into the expression x4+9x3+35x2−x+4{x^4} + 9{x^3} + 35{x^2} - x + 4 and simplifying results in -160.
269 Medium
269 of 337
Application +4 | -1
If ∣u∣<1,∣v∣<1|u| < 1,|v| < 1 and z=u−v1−uvz=\frac{u-v}{1-u v}, a lower bound of ∣z∣|z| is
View answer and explanation
Correct option: C
∥u∣−∣v∥1+∣u∣∣v∣\frac{\|u|-| v\|}{1+|u||v|}
Option C (∥u∣−∣v∥1+∣u∣∣v∣\frac{{\|u|-|v\|}}{{1+|u||v|}}) is correct because by using properties of complex numbers and the reverse triangle inequality, we can establish the lower bound for |z|.
270 Medium
270 of 337
Application +4 | -1
If ∣z+4i∣≤3|z+4 i| \leq 3 then maximum and minimum value of ∣z+3∣|z+3| is -
View answer and explanation
Correct option: A
8,2
Option A (8,2) is correct because the maximum and minimum values of ∣z+3∣|z+3| are found by using the triangle inequality on ∣(z+4i)+(3−4i)∣|(z+4i)+(3-4i)|.
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Review COMPLEX NUMBERS in your textbook or class notes, then use this set to check your understanding.

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